A single vertical slit of width 30um is located 2. 30m from a screen. The intens

Question

A single vertical slit of width 30um is located 2. 30m from a screen. The intensity pattern below is observed. The distances on the screen can be measured using the provided scale with a separation between the adjacent ticks of 1cm. What is the wavelength of the light? Answer: What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 505nm light? The index of refraction of the coating material is 1. 43 and the index of the glass is 1. 60. Answer:

Explanation / Answer

4. Distance of the n = 1 maxima from the central bright fringe = 6 cm

Wavelenth = (0.06)(30E-6)/(2.3) = 782.61 nm

5. The phase changes on reflection from both front and back surfaces cancel out so we can ignore them. The minimum path difference for destructive interference = ?/2 = 252.5 nm

The thickness that would give this pd is given by pd =

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